Question

Rewrite the given boundary value problem as an equivalent boundary value problem for a first order system. Your rewritten boundary value problem should have the form of equation (1): $$ \begin{aligned} &\mathbf{y}^{\prime}=A(t) \mathbf{y}+\mathbf{g}(t), \quad a

Short answer

Question: Rewrite the second-order boundary value problem as a first-order system in matrix form: $$ (ty''(t))' + e^t y(t) = 2, \quad y'(1) = -3, \quad y(2) = 1 $$ Answer: The first-order system for the given boundary value problem is: $$ \begin{aligned} &\mathbf{y}^{\prime}(t)=A(t) \mathbf{y}(t)+\mathbf{g}(t), \quad 1<t<2 \\\ &P^{[1]} \mathbf{y}(1)+P^{[2]} \mathbf{y}(2)=\alpha \end{aligned} $$ where \(A(t) = \begin{bmatrix} 0 & 1\\ -e^{t} & -\frac{1}{t}\end{bmatrix}\), \(P^{[1]} = \begin{bmatrix}0 & 1\end{bmatrix}\), \(P^{[2]} = \begin{bmatrix}1 & 0\end{bmatrix}\), \(\mathbf{g}(t) = \begin{bmatrix} 0 \\ \frac{2}{t}\end{bmatrix}\), and \(\alpha=\begin{bmatrix}1\end{bmatrix}\).

Step-by-step solution

Rewrite second-order equation as two first-order equations

Let \(y_1(t) = y(t)\) and \(y_2(t) = y'(t)\). Then we have: $$ \begin{aligned} y_1'(t) &= y_2(t) \\ (ty_2'(t))' + e^t y_1(t) &= 2 \\ \end{aligned} $$ Taking the derivative of the first equation with respect to t, we get: $$ \begin{aligned} y_2'(t) &= \frac{1}{t}[(ty_2'(t))'] - \frac{y_2(t)}{t} \\ \end{aligned} $$ Substitute the second equation into it: $$ \begin{aligned} y_2'(t) &= \frac{1}{t}[2 - e^t y_1(t)] - \frac{y_2(t)}{t} \\ \end{aligned} $$ Now we have two first-order equations: $$ \begin{aligned} y_1'(t) &= y_2(t) \\ y_2'(t) &= \frac{1}{t}[2 - e^t y_1(t)] - \frac{y_2(t)}{t} \\ \end{aligned} $$

Write the first-order equations in matrix form

Let \(\mathbf{y}(t)=\begin{bmatrix}y_1(t) \\ y_2(t)\end{bmatrix}\). Rewrite the system of first-order equations in the matrix form: $$ \begin{aligned} \mathbf{y}'(t) &= \begin{bmatrix} 0 & 1 \\ -e^{t} & -\frac{1}{t} \end{bmatrix} \mathbf{y}(t) + \begin{bmatrix} 0 \\ \frac{2}{t} \end{bmatrix} \end{aligned} $$ So we have \(A(t) = \begin{bmatrix} 0 & 1\\ -e^{t} & -\frac{1}{t}\end{bmatrix}\) and \(\mathbf{g}(t) = \begin{bmatrix} 0 \\ \frac{2}{t}\end{bmatrix}\).

Rewrite boundary conditions for the new system

The boundary conditions for the given second-order problem are: \(y'(1) = y_2(1) = -3\) and \(y(2) = y_1(2) = 1\). Rewrite these conditions for the new system: $$ \begin{aligned} P^{[1]} \mathbf{y}(1) + P^{[2]} \mathbf{y}(2) =\alpha \end{aligned} $$ Using the boundary conditions, we can find \(P^{[1]}\), \(P^{[2]}\), and \(\alpha\): $$ \begin{aligned} \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} y_1(1)\\ y_2(1) \end{bmatrix} + \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} y_1(2)\\ y_2(2) \end{bmatrix} = \begin{bmatrix} 1 \end{bmatrix} \end{aligned} $$ So, \(P^{[1]} = \begin{bmatrix}0 & 1\end{bmatrix}\), \(P^{[2]} = \begin{bmatrix}1 & 0\end{bmatrix}\), \(\alpha=\begin{bmatrix}1\end{bmatrix}\). The first-order system for the given boundary value problem is rewritten as: $$ \begin{aligned} &\mathbf{y}^{\prime}(t)=A(t) \mathbf{y}(t)+\mathbf{g}(t), \quad 1<t<2 \\\ &P^{[1]} \mathbf{y}(1)+P^{[2]} \mathbf{y}(2)=\alpha \end{aligned} $$ where \(A(t) = \begin{bmatrix} 0 & 1\\ -e^{t} & -\frac{1}{t}\end{bmatrix}\), \(P^{[1]} = \begin{bmatrix}0 & 1\end{bmatrix}\), \(P^{[2]} = \begin{bmatrix}1 & 0\end{bmatrix}\), \(\mathbf{g}(t) = \begin{bmatrix} 0 \\ \frac{2}{t}\end{bmatrix}\) and \(\alpha=\begin{bmatrix}1\end{bmatrix}\).

Key concepts

First Order System

When dealing with differential equations, especially those of higher order, it can be extremely beneficial to reduce them to a first order system. This approach simplifies the problem, making it more manageable and often easier to analyze or solve numerically. In the context of our exercise, the original second-order differential equation is transformed into a system that consists solely of first order differential equations.

In general, a first order system can be represented as \(\mathbf{y}' = A(t) \mathbf{y} + \mathbf{g}(t)\), where \(\mathbf{y}\) is a vector of unknown functions, \(A(t)\) is a matrix that can depend on the independent variable \(t\), and \(\mathbf{g}(t)\) is a vector function of \(t\). To convert a higher order differential equation to a first order system, we introduce new variables to represent each derivative up to the \(n-1\)st order for an \(n\)th order differential equation.

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are crucial in modeling the behavior of various natural and engineered systems. In our exercise, we encounter a boundary value problem that involves a differential equation and specific values, or 'boundary conditions', at the interval's endpoints.

The initial higher order differential equation in this case is \(\left(t y'\right)' + e^t y = 2\), with the boundary conditions \(y'(1) = -3\) and \(y(2) = 1\). The beauty of differential equations lies in their ability to encapsulate a physical, biological, or economic process into a mathematical framework that can then be analyzed and solved to predict the behavior of the system in question.

Matrix Form Differential Equations

Matrix form differential equations provide a structured way to write systems of first order differential equations. In this format, the system of equations is compactly expressed using a single matrix multiplication, which makes it particularly useful for statistical software and numerical methods. In the exercise, we express the system as \(\mathbf{y}'(t) = A(t)\mathbf{y}(t) + \mathbf{g}(t)\), where \(A(t)\) is the matrix containing the coefficients of the system and \(\mathbf{g}(t)\) is the vector function representing the non-homogeneous part of the system.

By converting to this form, we are able to leverage powerful linear algebra techniques and computational tools to find solutions, analyze stability, and understand the dynamics of the system. This approach is central to modern applied mathematics, and is particularly important in the fields of control theory, signal processing, and many areas of engineering.